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Copyright © 2012 Pearson Education, Inc. Slide 7- 1 5.3 Polynomial Equations and Factoring ■ ■ ■ Graphical Solutions The Principle of Zero Products Terms with Common Factors ■ Factoring by Grouping ■ Factoring and Equations Copyright © 2012 Pearson Education, Inc. Graphical Solutions Whenever two polynomials are set equal to each other, the result is a polynomial equation. In this section we learn how to solve polynomial equations both graphically and algebraically by factoring. Copyright © 2012 Pearson Education, Inc. Slide 5- 3 Example Solve: x2 = 4x. Solution Intersect Method: We can find the real-number solutions of a polynomial equation by finding the points of intersection of two graphs. Copyright © 2012 Pearson Education, Inc. Slide 5- 4 continued Zero Method: Rewrite the equation so that one side is 0 and then find the x-intercepts of one graph or the zeros of the function. Rewrite the equation so that one side is 0: x2 = 4x x2 – 4x = 0 Copyright © 2012 Pearson Education, Inc. Slide 5- 5 Zeros and Roots The x-values for which a function f(x) is 0 are called the zeros of the function. The x-values for which an equation such as f(x) = 0 is true are called the roots of the equation. Copyright © 2012 Pearson Education, Inc. Slide 5- 6 Example Find the zeros of the function given by f(x) = x3 – 2x2 – 5x + 6. Solution Graph the equation, choosing a window that shows the x-intercepts of the graph. This may require several attempts. To find the zeros use the ZERO option from the CALC menu. Copyright © 2012 Pearson Education, Inc. Slide 5- 7 continued f(x) = x3 – 2x2 – 5x + 6 To find the zeros use the ZERO option from the CALC menu. Use the same procedure for the other two zeros. The zeros are 2, 1 and 3. Copyright © 2012 Pearson Education, Inc. Slide 5- 8 The Principle of Zero Products The Principle of Zero Products For any real numbers a and b: If ab = 0, then a = 0 or b = 0. If a = 0 or b = 0, then ab = 0. Copyright © 2012 Pearson Education, Inc. Slide 5- 9 Example Solve: (x – 4)(x + 3) = 0. Solution According to the principle of zero (4, 0) products, at least one factor (3, 0) must be 0. x–4=0 or x + 3 = 0 x=4 or x = 3 For 4: For 3: (x – 4)(x + 3) = 0 (x – 4)(x + 3) = 0 (4 – 4)(4 + 3) = 0 (3 – 4)(3 + 3) = 0 0(7) = 0 0(7) = 0 0 = 0 TRUE 0 = 0 TRUE Copyright © 2012 Pearson Education, Inc. Slide 5- 10 Terms with Common Factors When factoring a polynomial, we look for factors common to every term and then use the distributive law. Multiply Factor 4x(x2 + 3x 4) = 4xx2 + 4x3x 4x4 = 4x3 + 12x2 16x 4x3 + 12x2 16x = 4xx2 + 4x3x 4x4 = 4x(x2 + 3x 4) Copyright © 2012 Pearson Education, Inc. Slide 5- 11 Example Factor: 28x6 + 32x3. Solution The prime factorization of 28x6 is 227xxxxxx The prime factorization of 32x3 is 22222xxx The largest common factor is 2 2 x x x or 4x3. 28x6 + 32x3 = 4x3 7x3 + 4x3 8 = 4x3(7x3 + 8) Copyright © 2012 Pearson Education, Inc. Slide 5- 12 Example Factor: 12x5 21x4 + 24x3 Solution The prime factorization of 12x5 is 223xxxxx The prime factorization of 21x4 is 37xxxx The prime factorization of 24x3 is 2223xxx The largest common factor is 3 x x x or 3x3. 12x5 21x4 + 24x3 = 3x3 4x2 3x3 7x + 3x3 8 = 3x3(4x2 7x + 8) Copyright © 2012 Pearson Education, Inc. Slide 5- 13 Factoring by Grouping Sometimes algebraic expressions contain a common factor with two or more terms. Example Factor x2(x + 2) + 3(x + 2). Solution The binomial (x + 2) is a factor of both x2(x + 2) and 3(x + 2). Thus, x + 2 is a common factor. x2(x + 2) + 3(x + 2) = (x + 2)x2 + (x + 2)3 = (x + 2)(x2 + 3) Copyright © 2012 Pearson Education, Inc. Slide 5- 14 Example Write an equivalent expression by factoring. a) 3x3 + 9x2 + x + 3 b) 9x4 + 6x 27x3 18 Solution a) 3x3 + 9x2 + x + 3 = (3x3 + 9x2) + (x + 3) = 3x2(x + 3) + 1(x + 3) = (x + 3)(3x2 + 1) Don’t forget to include the 1. Copyright © 2012 Pearson Education, Inc. Slide 5- 15 Example continued b) 9x4 + 6x 27x3 18 = (9x4 + 6x) + (27x3 18) = 3x(3x3 + 2) + (9)(3x3 + 2) = (3x3 + 2)(3x 9) = (3x3 + 2)3(x 3) = 3(3x3 + 2)(x 3) Copyright © 2012 Pearson Education, Inc. Slide 5- 16 Factoring and Equations Example Solve: 7x2 = 35x. Solution Use the principle of zero products if there is a 0 on one side of the equation and the other side is in factored form. 7x2 = 35x 7x2 – 35x = 0 Subtracting 35x. One side is now 0. 7x(x – 5) = 0 Factoring 7x = 0 or x – 5 = 0 Use the principle of zero products x=0 or x=5 We check by substitution or graphically. Copyright © 2012 Pearson Education, Inc. Slide 5- 17 To Use the Principle of Zero Products 1. Write an equivalent equation with 0 on one side, using the addition principle. 2. Factor the nonzero side of the equation. 3. Set each factor that is not a constant equal to 0. 4. Solve the resulting equations. Copyright © 2012 Pearson Education, Inc. Slide 5- 18